# Number of modes in incoming light

1. Jul 23, 2006

### octol

Hello all,

if I have incoming radiation from a blackbody source filtered to a bandwidth of 0.1 nm and centered at a wavelength of 500 nm, why is the number of modes in this light not equal to the density of photon states times the bandwidth?

I.e why isn't it
$$\text{number of modes} = g(\omega) * \delta \omega = \frac{\omega^2}{\pi^2 c^3} \delta \omega$$
where
$$\omega = \frac{2 \pi c}{500 nm}$$
and
$$\delta \omega = \frac{2 \pi c}{0.1 nm}$$ ?

If anyone knows why I'd be very thankful for an explanation.

Best regards,
Jon

2. Jul 23, 2006

### Gokul43201

Staff Emeritus
You're omitting the $exp(\hbar \omega/kT ) - 1$ factor in the denominator, which comes from the occupation fraction of a Bose gas, if you are interested in the number of occupied states and not just the number of available states.

3. Jul 26, 2006

### octol

Yes but I thought that adding that factor would give me the number of photons? And that the there are more than one photon for every mode (frequency). Or have I missunderstood the question? I thought that "mode" refers to the number of available frequency states?

I can't say I've fully grasped the difference between "state", "mode" etc yet